Answer:
The radius of the slave piston = 14.10 cm
Explanation:
The formula to apply here is that for pressure where ;
Pressure = Force / Area
F₁ /A₁ =F₂/A₂ where ; F₁=200 N , A₁= 25 cm² , F₂ =5000 N ,A₂ =?
Apply values in the equation as ;
200/ 25 = 5000/ A₂
A₂ = 625 cm²
Formula for area is ;
A = πr²
625 = π * r²
625/π = r²
√625/π =r
14.10 cm = r
show your work. john uses a 25N force to push a boulder off a cliff that is 312m tall. What is the work done on the boulder?
Answer:7800
work=force x distance
Force in Newtons
Distance in Meters
Work in Joules
WRITE A PARAGRAPH ABOUT PRESSURE
Answer:
pressure is a horrible thing to go through, it can lead to many bad side effects, like burn out and it can also lead to high blood pressure, head aches, heart problems, depression, anxiety, and many more damaging effects. Pressure can destroy someones mental health if not dealed worth properly
I hope this is okay! I'm not sure what pressure you meant
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
To learn more about the spring force refer to the link;
https://brainly.com/question/4291098
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1. While riding a chairlift, a 55-kg skier is raised a vertical distance of 370 m. What is the total change in the skier's gravitational potential energy
Answer:
199,430Joules
Explanation:
Gravitational potential energy is the energy possessed by a falling object due to virtue of its position.
gravitational potential energy = mass * acceleration due to gravity * height
gravitational potential energy = 55 * 9.8 * 370
gravitational potential energy = 539 * 370
gravitational potential energy = 199,430
Hence the total change in the skier's gravitational potential energy is 199,430Joules
ANSWER:
99,430 Joules (in multiple choice question sometimes appear as 200,000J)
Explanation:
Gravitational potential energy is the energy possessed by a falling object due to virtue of its position.
gravitational potential energy = mass * acceleration due to gravity * height
gravitational potential energy = 55 * 9.8 * 370
gravitational potential energy = 539 * 370
gravitational potential energy = 199,430
Hence the total change in the skier's gravitational potential energy is 199,430 Joules (or 200,00J)
please heart!VVV or stars :D vvv
NEED AWNSER NOW! WILL MARK BRAINLY! Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through?
index of reflection
index of refraction
angle of reflection
angle of incidence
Answer:
Index of refraction.
Answer:
index of refraction
Explanation:
I just took the k12 quiz.
Jerry is pushing a 50-kg box across a moth floor with an acceleration of 0.6 m/s2. What force is he applying to the box? *
83.3 N
0.012 N
0
30 N
Answer:
30 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 50 × 0.6
We have the final answer as
30 NHope this helps you
Can u anser 5,6 on the picture
Answer: Number 6 is Periods
Explanation:
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
A 250-kg moose stands in the middle of the railroad tracks in Sweden, frozen by the lights of an oncoming 10,000kg train traveling at 20m/s. Even though the engineer attempted in vain to slow the train down in time to avoid hitting the moose, the moose rides down the remaining track sitting on the train’s cowcatcher. What is the final velocity of the train and moose after the collision?
(Momentum & Impulse)
Answer:
The final velocity of the train and the moose after collision is approximately 19.51 m/s
Explanation:
The given mass of the moose, m₁ = 250 kg
The velocity of the moose, v₁ = 0
The mass of the oncoming train, m₂ = 10,000 kg
The velocity of the train, v₂ = 20 m/s
The velocity of the moose and the train after collision = v₃
By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃
Therefore, by substitution, we have;
250×0 + 10,000× 20 = (10,000 + 250) × v₃
200,000 = 10,250 × v₃
v₃ = 200,000/10,250 ≈ 19.51 m/s
The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s
Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente hacia adelante para que cuando ésta toque la superficie del agua haga varios "saltos" sobre el agua. Durante un juego de estos, un desocupado nota que una de las piedras que arroja se demora 0,4 s en tocar la superficie del agua y la toca a 2,5 m de la orilla del lago, desde donde fue lanzada. Encuentre: a) La altura de la que fue lanzada la piedra. b) La velocidad con la que fue lanzada.
Answer:
a) La piedra es lanzada desde una altura de 0,785 metros.
b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
Explanation:
a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:
[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Donde:
[tex]y[/tex] - Altura final, medida en metros.
[tex]y_{o}[/tex] - Altura inicial, medida en metros.
[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.
Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:
[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]
[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]
[tex]y_{o} = 0,785\,m[/tex]
La piedra es lanzada desde una altura de 0,785 metros.
b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:
[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)
Donde:
[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.
[tex]t[/tex] - Tiempo, medido en segundos.
Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:
[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]
[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]
La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
In a place covered by shadow of cloud sun cannot be seen . Explain with reasons .
Answer:
Because even though our eyes have a huge dynamic range (ability to pick out details in sharply lit and lesser lit areas simultaneously) than any camera, there's a limit.
When there's strong sunlight, your pupils contract and let less light in, which makes the shadows look darker.
When it's cloudy, your pupils widen and let more light in, which makes the shadows look less dark.
Do some experiments with a camera and you'll soon get the hang of it.
NOTE: Also test HDR (high dynamic range) photography, where the camera takes three or more pictures in quick succession, with different exposure settings, and combines them to get the most detail of both bright and dark areas. The result is more or less what we percieve.
A golf ball is sitting on a tee. The ball is struck with a golf club and flies
through the air. How does the force on the club compare with the force on the
ball when momentum is transferred between the club and ball?
Answer:
c i kn now it is
Explanation:
Which electron dot diagram shows the bonding between 2 chlorine atoms?2 dots then C l with 2 dots above and 1 dot below then 2 dots then 2 dots then C l with 2 dots above and 1 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 2 dots then C l with 2 dots above and 2 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 1 dot then C l with 2 dots above and 2 dots below then 2 dots.2 dots then C l with 2 dots above and 1 dot below then 3 dots then 3 dots then C l with 2 dots above and 1 dot below then 2 dots.
Answer:
It is B
Explanation:
Answer: 2nd answer
Explanation: took exam
The diagram shows a model of an atom. Who first proposed this model?
A. Bohr
B. Thomson
C. Rutherford
ОО
D. Dalton
A.Bohr
His model postulated the existence of energy levels or shells of electrons. Electrons could only be found in these specific energy levels; in other words, their energy was quantised, and couldn’t take just any value. Electrons could move between these energy levels but had to do so by either absorbing or emitting energy.
A. Bohr!
This answer is correct because I read the information.
Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero
Answer:
The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.
What muscle movement do we use to close a joint.
Answer: Your using your skeletal muscle
Explanation:
A fish swimming at a rate of .6 m/s notices a huge shark. Three seconds later, the fish is swimming at a speed of 3 m/s. What is the fish's acceleration?
0.8 m/s/s
-0.8 m/s/s
12.5 m/s/s
-12.5 m/s/s
Answer:
C
Explanation:
???
i think
9. A student notices that wearing darker colors in sunlight makes him feel warmer, so he decides to conduct an experiment. He takes five pieces of different
colored cloth and wraps
each one around a water bottle. He then places all five bottles in direct sunlight and measures the temperature of the water in each bottle an hour later
What is the dependent variable in this experiment?
O the time he leaves it in the sunlight
O the amount of water in each bottle
O the color of the cloth
O the temperature of the water
Answer: 4
Explanation:
The dependent variable is the temperature of the water.
A 75kg bicyclist (including the bicycle), initially at rest at the top of a hill, coasts down the hill, reaching a speed of 14.6m/s at the bottom of the hill. The distance and height of the hill are shown. Neglect any friction impeding the motion and the rotational energy of the wheels. List the energy types at the initial and final time and whether work and loss (due to non-conservative forces) occur as well as the corresponding amounts of energy.
The energy type at the initial time is potential energy and the energy at the final time or position is kinetic energy.
What is the law of conservation of energy?The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.
Based on the law of conservation of mechanical energy, the formula for the change in the kinetic energy and the potential energy of the bicyclist is given as;
K.Ei + P.Ei = K.Ef + P.Ef
where;
K.Ei is the initial kinetic energy of the bicyclistK.Ef is the final kinetic energy of the bicyclistP.Ei is the initial potential energy of the bicyclistP.Ef is the final potential energy of the bicyclistThe kinetic energy of the bicyclist increases with increase in the velocity of the bicyclist while the potential energy increases with increase in the height of the bicyclist.
At the initial position when the bicyclist is at rest, the kinetic energy is zero, so the only energy at the initial position is potential energy because the height is maximum.
In addition, at the final position, the velocity of the bicyclist is maximum and the height is zero, so the only energy at the final position is kinetic energy.
Learn more about conservation of energy here: https://brainly.com/question/166559
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A friend is coming to Tim’s house to study after school. What directions would Tim give for reaching his house from the entrance of the school?
(I WILL GIVE BRAINLIEST)
Answer:
go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school
How much would a pair of 0.5 kg shoes weigh on Earth? (Include units in
your answer) *
Answer:
1.1 lbs
Explanation:
To convert kg to lbs you multiply kilograms by 2.2. So 0.5kg × 2.2 equals to 1.1 lbs
draw position time graph when speed is increasing
Explanation:
We need to draw position-time graph when the speed is increasing.
The slope of position-time graph gives the speed of an object.
Position means distance covered.
When the speed of an object is increasing with time. It means it is moving with increasing speed.
The attached figure shows the position -time graph when speed is increasing.
A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which water hits a wall at 8 m away from the hose (consider that acceleration due to gravity =9.8 m/s2)
Answer:
The water hits the wall at a height of 5.38 m
Explanation:
Projectile Motion
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
The object describes a parabolic path given by the equation:
[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]
Where:
y = vertical displacement
x = horizontal displacement
θ = Elevation angle
vo = Initial speed
The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.
The height at which the water hits a wall located at x=8 m from the hose is:
[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]
Calculating:
y = 5.38 m
The water hits the wall at a height of 5.38 m
Give one example where friction is useful. Can someone feed me with correct answer pls c:
Answer:
Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.
Explanation:
When two ocean plates come together, one ocean plate __________________
under the other, causing a chain of ________________ __________________
to form.
Answer:
A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.
Explanation:
help please asap due 20 minutes please help me
An object that falls and accelerates solely as a result of gravity is said to be in
(2 points)
A. terminal velocity
B. free fall
C. air resistance
D. terminal acceleration
A 0.15 kg ball is moving with a velocity of
35 m/s. Find the momentum of the ball.
Answer:
5.25 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 0.15 × 35
We have the final answer as
5.25 kg.m/sHope this helps you
5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley to a 0.0150 kg mass.
Draw a free body diagram showing all the forces acting on each object
Calculate the acceleration of the masses
Calculate the tension force in the cord
How long will it take the block to get to go 1.2 m to the edge of the table.
What will the velocity be as soon as it gets to the edge?
Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
Answer:
There's no answer I'm just taking points like you did me, so thank you for your points I'll put them to good use ;)
13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters
Answer:
B. 1200
Explanation:
60 sec in one min in 2 min there will be 120 sec. 10x120=1200